Question

If $\omega$ is the imaginary cube root of the unity, then ${{(1+\omega -{{\omega }^{2}})}^{7}}$ equal
$A)128\omega$
$B)-128\omega$
$c)128{{\omega }^{2}}$
$D)-128{{\omega }^{2}}$

Answer 1

To solve this question we need to know the concept of imaginary numbers. Imaginary number is basically represented as $i$ which is iota. $\omega$ is an imaginary cube root of unity. To solve the question, some of the formulas required are ${{w}^{3}}=1$ and $1+w+{{w}^{2}}=0$ .

Complete step by step answer:
The question ask us to find the value of ${{(1+\omega -{{\omega }^{2}})}^{7}}$ when $\omega$ is the imaginary cube root of the unity. The first step is to consider the fact regarding the formula ${{w}^{3}}=1$ and $1+w+{{w}^{2}}=0$ . Now we will start with the expression we are given to solve which is :
$\Rightarrow {{(1+\omega -{{\omega }^{2}})}^{7}}$
We will substitute the formula of the omega, $\omega$ which says $1+w+{{w}^{2}}=0$. On rearranging the formula we get: $\Rightarrow 1+w=-{{w}^{2}}$
Substituting the above formula we get:
$\Rightarrow {{(-{{\omega }^{2}}-{{\omega }^{2}})}^{7}}$
On calculating the above expression we get:
$\Rightarrow {{(-2{{\omega }^{2}})}^{7}}$
We know that the negative number having odd power gives negative as the result. So on calculating the expression given above the answer will be negative. On solving further we get:
$\Rightarrow -128{{\omega }^{14}}$
Now it is the time to apply the other formula which says ${{w}^{3}}=1$. So we will write the power $14$ in expanded form where $3$ will be the divisor. So $14$ will be written as:
$\Rightarrow -128{{\omega }^{3\times 4+2}}$
From the above expression we know that the ${{\omega }^{3n}}$ will always be $1$ . So on writing it in the expanded form we get:
$\Rightarrow -128{{\omega }^{3\times 4}}{{\omega }^{2}}$
$\Rightarrow -128\times 1\times {{\omega }^{2}}$
$\Rightarrow -128{{\omega }^{2}}$
$\therefore$ The value of ${{(1+\omega -{{\omega }^{2}})}^{7}}$ when $\omega$ is the imaginary cube root of the unity is $D)-128{{\omega }^{2}}$.

So, the correct answer is “Option D”.

Note: Imaginary cube root of unity which means $1$ is $1,\omega$ and ${{\omega }^{2}}$ . Here $\omega$ and ${{\omega }^{2}}$are the conjugates of an imaginary number $-\dfrac{1}{2}+i\dfrac{\sqrt{3}}{2}$ . So the three roots of unity are $1,-\dfrac{1}{2}+i\dfrac{\sqrt{3}}{2}$ and $-\dfrac{1}{2}-i\dfrac{\sqrt{3}}{2}$, which means one real and two imaginary roots.